0
$\begingroup$

Let $X$ be a topological space and $A$ be a non-empty subset of $X$. Then one can conclude that if $X\setminus A$ is nowhere dense in $X$, $A$ is dense in $X$,

Is the above statement true in general? I know if $A$ is open then the result is true, but I am not sure otherwise.

$\endgroup$
4
$\begingroup$

No, that is not true in general.

Consider $X = \mathbb{R}$. Clearly $A = \mathbb{Q}$ is dense in $\mathbb{R}$, but $\mathbb{R} \setminus \mathbb{Q}$ is also known to be dense as well.

$\endgroup$
  • $\begingroup$ Thank you so much $\endgroup$ – Golam biswas Jun 25 '18 at 17:40
1
$\begingroup$

No, as others pointed out already. Sets can be dense and co-dense, like the rationals and the irrationals in the real numbers.

But $A$ is nowhere dense iff $X\setminus \overline{A}$ is dense. So something close to it is true.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.